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[GreatJustice]

Alright, two things I'm having some issues with.

First, if I have a prime and two integers x and y s.t p|x^2 + y^2 but p doesn't divide x or y, how would I go about proving that there is a k s.t k^2 is congruent to -1 (mod p)? I feel like I'm missing something quite obvious.

Also, how would I prove whether there is or isn't a sequence of five natural numbers s.t none are square free?

[Helgoland]

-snip-

Thanks - I actually had that lecture open in another tab Hadn't noticed that example though. Now if only my Topology Intro or Topology 1 had treated at least one third of the concepts in there...

E: Couldn't you just start with k=(x+y+i) for some i \in N, with k^2= x^2 + y^2 + 2xy + 2i(x+y) + i^2? Then k^2 is congruent to 2xy + 2i(x+y) + i^2, which is just some polynomial of degree two and ought to hit any number you want as i varies, if I remember my algebra right.

Latter one: Just look at the multiples of 4=2^2: They occur every four integers, so every sequence of five contains at least one...

[ZetaX]

Now if only my Topology Intro or Topology 1 had treated at least one third of the concepts in there...

I would say that a lot of it is more akin to homological algebra, which a lot of topology courses seem to skip over unless they cannot longer avoid it (e.g. Hatcher's textbook does this all the time). For beginner's courses this is maybe fine, but might still come back to bite you later.

First, if I have a prime and two integers x and y s.t p|x^2 + y^2 but p doesn't divide x or y, how would I go about proving that there is a k s.t k^2 is congruent to -1 (mod p)? I feel like I'm missing something quite obvious.

Essentially: just take k = x/y. As an element of a finite field this is directly defined; more classically (but actually the same), you would use xy^(-1), where z=y^(-1) is any integer inverse to y mod p, i.e. yz = 1 mod p.

Also, how would I prove whether there is or isn't a sequence of five natural numbers s.t none are square free?

You can do it for k instead of 5 as well: Use the Chinese Remainder Theorem to find a sequence a, a+1, ..., a+k-1 such that (p_i)² divides a+i (i.e. a = -i mod (p_i)²), where {p_i | i} is any list of distinct primes.

[Furtuka]

Hi I'm having a bit of a problem with one question that asking me to integrate 1 by dy dx over 1 to Sqrt(1-x^2) and 0 to 1. I'm lost at how to handle the second integral in it. The book answers say I should recognize it's a quarter circle area and go from there, but doesn't elaborate on why I should, or what to do from there.

Also how should I write an integral that has the domain of Absolute Value of X ≤ 1?

[crazysheep]

Have you covered changing coordinates in your classes yet? The hint about the quarter circle certainly nudges me in that direction, so you want to change from the x-y coordinates to r-theta coordinates.

Not sure about the absolute value domain question though.

[Zrk2]

Have you covered changing coordinates in your classes yet? The hint about the quarter circle certainly nudges me in that direction, so you want to change from the x-y coordinates to r-theta coordinates.

Not sure about the absolute value domain question though.

This is my suggestion, too. Alternatively you can do the double integral, but it may get messy, although I haven't pulled out the paper required to check.

[RedWarrior0]

For the first question, if I'm understanding it correctly, you're being asked to evaluate this? In which case looking at the second integral, it makes a lot of sense to change coordinates.

For the second question, isn't it just Integral[f(x) dx] from -1 to 1, or am I thinking something else?

[Zrk2]

Nope. Sometimes you gotta do things the long way.

[Helgoland]

Or use Wolfram Alpha.

[Reelya]

it might be a little less painful to expand one term at a time and divide as you go along:

(x+y)(x+z)(y+z)(x+y+z)/xyz=0

split by the (x+y+z) term

x * (x+y)(x+z)(y+z)/xyz + y * (x+y)(x+z)(y+z)/xyz + z * (x+y)(x+z)(y+z) / xyz = 0

here you can already start canceling out top/bottom factors:

(x+y)(x+z)(y+z)/yz + (x+y)(x+z)(y+z)/xz + (x+y)(x+z)(y+z) / xy = 0

now split each term by the one that lets you cancel the most fractions:

y(x+y)(x+z)/yz + z(x+y)(x+z)/yz + x(x+y)(y+z)/xz + z(x+y)(y+z)/xz + x(x+z)(y+z) / xy + y(x+z)(y+z) / xy = 0

another round of cancelling:

(x+y)(x+z)/z + (x+y)(x+z)/y + (x+y)(y+z)/z + (x+y)(y+z)/x + (x+z)(y+z) / y + (x+z)(y+z) / x = 0

Now you only have to expand quadratic expressions, which is much simpler.

[RedWarrior0]

Thanks Reelya. As for using Wolfram Alpha, it sadly does not simplify the polynomial to quite what I want.

Now, to do this on a polynomial with 260 terms.

For that, you leave the bounds of Mathematics and enter the realm of Computer Science.

[frostshotgg]

Thanks Reelya. As for using Wolfram Alpha, it sadly does not simplify the polynomial to quite what I want.

Now, to do this on a polynomial with 260 terms.

You have C++ do that for you.

[MagmaMcFry]

Why do you need to do that anyway? I can't think of anything that you can do with polynomials in expanded form, but not in product form.

[frostshotgg]

I would say do it by hand for 4 or 5 terms and see if you can find something.

[ZetaX]

So, here's some math that should be straightforward- polynomials.

Say I have the polynomial product (x+y)(x+z)(y+z)(x+y+z)=0. Now, I know that if you divide the entire thing by xyz, you are left with (x+y)²/z + (x+z)²/y + (y+z)²/x + 4x + 4y + 4z = 0.

My question is: other than fully expanding the entire thing out, is there any other way to get from one equation to the other?

In the following I am assuming that you want to show that those two equations are equivalent. You have several options:

a) Just expand. The long, tedious and boring way, but requires no thought at all. Best done when sleepy or if a slave is available.

b) You already know a neat factorisation of one equation. The second one has the same degree after multiplaying with xyz (you can see that without actually multiplying). Thus for them to be the same equation one only needs to check that the second one vanishes if x+y=0, y+z=0, z+x=0 or x+y+z=0. As the equations are symmetric (i.e. interchanging the variables does  not change either of them), we only would need to check vanishing if z+x=0 and x+y+z=0. So substitute z=-x and check for the second equation to be satisfied; then do the same with z=-x-y.
Still somewhat lengthy, but already shorter; has the disadvantage of needing the first one to be neatly factored which won't always be so.

c) Use the weak version the combinatorial nullstellensatz (Theorem 1.2 in http://www.tau.ac.il/~nogaa/PDFS/null2.pdf). In other words: polynomials are equal if they give the same result for enough choices of values for x,y,z.
In this case: multiply the second equation by xyz to get a polynomial. Then check if both polynomials from the two equations give the same values when plugging in sufficiently many values. Sufficiently many are e.g. each of the 5³=125 possible combinations of putting each of x,y,z to one of -2,-1,0,1,2. You would (by fully invoking that theorem and symmetry) actually only need to check 5+8+12 = 25 triples:
y=z=0, x from -2,-1,0,1,2
z=0, y from 0,1, x from -1,0,1,2
z from 0,1, y from 0,1, x from -1,0,1
A weird way to do it. Some slave or too much spare time recommended, but actually not that much work (those 0's cancel a lot).